Cauchy-Schwartz Inequality Proof

1. The problem statement, all variables and given/known data
Show that |<v|w>|^2 ≤ <v|v><w|w>
for any |v>,|w> ∈ ℂ^2

2. Relevant equations

3. The attempt at a solution

The Cauchy-Schwartz inequality is extremely relevant for the math/physics that I am interested in.
I feel like I have a very good proof here, but I am interested in a few things here.
1) Is my proof correct in the right context (Complex numbers here)
2) What is your favorite proof of the Cauchy Schwartz Inequality? (I want to master this one)

Proof:
Case 1: We have to show that if |v> or |w> = 0, then the sides are equal to 0. Which is somewhat trivial to prove by simply plugging in the values and reaching the conclusion using inner space axioms.

Case 2: Where |v> and |w> are not equal to 0.
Here, I let:
a = <v|v>
b = 2<v|w>
c = <w|w>

1. The problem statement, all variables and given/known data
Show that |<v|w>|^2 ≤ <v|v><w|w>
for any |v>,|w> ∈ ℂ^2

2. Relevant equations

3. The attempt at a solution

The Cauchy-Schwartz inequality is extremely relevant for the math/physics that I am interested in.
I feel like I have a very good proof here, but I am interested in a few things here.
1) Is my proof correct in the right context (Complex numbers here)
2) What is your favorite proof of the Cauchy Schwartz Inequality? (I want to master this one)

Proof:
Case 1: We have to show that if |v> or |w> = 0, then the sides are equal to 0. Which is somewhat trivial to prove by simply plugging in the values and reaching the conclusion using inner space axioms.

Case 2: Where |v> and |w> are not equal to 0.
Here, I let:
a = <v|v>
b = 2<v|w>
c = <w|w>

So, my one concern is that this may not be complete for complex inner products, but rather only for real numbers? Is there any cause for that concern?

You can answer that for yourself, by answering the following questions:
In line 1 of your development, is <tv + w | tv + w> real>? Is t real?
In line 2, are <v|v>, 2, <v|w> and <w|w> real?
So, in line 3 are a, b, c and t real?

So we see that in the part: 2<v|w>t where I am stating that b = 2<v|w>
This does NOT have to be real, it can be complex.

But the answer for this inequality to hold must be real.

So what we are really stating here is that b = 2<v|w> does not need to be real based on our definitions.

And therefore this implies that the discriminant proof stating that the polynomial either has no real roots or a repeated root is therefore junk, garbage, not true.

So this proof is entirely incorrect.

Right?

If your vector space is over the real field, you are done; your proof is correct and complete in that case. However, if it is over the complex field you have more work to do. In that case it is, indeed, important that t be allowed to assume complex values.

So I have to adjust this part.
I'm just no longer sure where to go? There must be a property of inner products I need to use.
Is there a property that will allow me to switch the components? So that I can make this work?

I must ask, why did you change r to x prior to taking the discriminat?
Was this to make it easier to visualize in quadratic form, or is there a more significant meaning?

Initially, I said ##r \geq 0##. I could have changed that to "for all real ##r##"), but I thought it might be less confusing to change ##r## to ##x## and allow both signs of ##x##. Anyway, who cares if we call it ##r## or ##x##? It is just a variable. The main issue is the region over which that variable is allowed to roam---which is ultimately the whole real line in the end.